Answer:
1367.7 g of ethylene glycol was added to the solution
Explanation:
In order to find out the mass of glycol we added, we apply the colligative property of lowering vapor pressure: ΔP = P° . Xm
ΔP = Vapor pressure of pure solvent (P°) - Vapor pressure of solution(P')
525.8 mmHg - 451 mmHg = 451 mmHg . Xm
74.8 mmHg / 451 mmHg = Xm → 0.166 (mole fraction of solute)
Xm = Mole fraction of solute / Moles of solute + Moles of solvent
We can determine the moles of solvent → 2000 g . 1 mol/18 g = 111.1 mol
(Notice we converted the 2kg of water to g)
0.166 = Moles of solute / Moles of solute + 111.1 moles of solvent
0.166 (Moles of solute + 111.1 moles of solvent) = Moles of solute
18.4 moles = Moles of solute - 0.166 moles of solute
18.4 = 0.834 moles of solute → Moles of solute = 18.4/0.834 = 22.06 moles
Let's convert the moles to mass → 62 g/mol . 22.06 mol = 1367.7 g
Hello!
The reaction that the graph represents is
A. Exothermic because Hrxn=-167 kJTo calculate Hrxn we apply the following equation:

Looking at the graph, and at the result of the calculations, we can see that the enthalpy of the products is
lower than the enthalpy of the reagents, because the sign is negative. That means that the reaction
releases energy in the form of heat and that the reaction is
exothermic.
Have a nice day!
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<span>Active transport runs counter to facilitated diffusion. In active transport, molecules move against the concentration gradients, running from areas of lower concentration to areas of higher concentration. This is where energy is used.</span>
Question 22: conclusion
Question 23: analyze